Compact Sets in Hausdorff Spaces are Closed

Last updated Nov 1, 2022

Let $X$ be a Hausdorff Topological Space. Then if $K \subset X$ is Compact, $K$ is Closed.

Let $x \in K^{C}$. Because Compact Sets are Separable from Points in a Hausdorff Space, there exists Open $V_{x} \subset X$ such that $x \in V_{x}$ and $V_{x} \cap K = \emptyset$. That is, $V_{x} \subset K^{C}$. Then, we can create Open $$V := \bigcup\limits_{x \in K^{C}}V_{x}$$ Since, for each $x \in K^{C}$, $V_{x} \subset K^{C}$, we see $V \subset K^{C}$. On the other hand, for each $x \in K^{C}, x \in V_{x}$, so $x \in V$. Thus $K^{C} \subset V$. Thus $K^{C} = V$. By definition of Closed, $K$ is Closed. $\blacksquare$